Chapter 8.7, Problem 2E

(a)

To determine

To Find: The explanation for the given series no the Taylor series of $f$ ‘centred at 1.

Answer to Problem 2E

The series is not the series of f that is centred at $x=1$ as the graph of the function f increases $f^{\prime }\left(1\right)>0$ .

Explanation of Solution

Given:

The given graph is shown in Figure 1

  

Figure 1

The given series is,

  $1.6-0.8\left(x-1\right)+0.4{\left(x-1\right)}^2-0.1{\left(x-1\right)}^3+\mathrm{...}$ .

Calculation:

Consider that the series is the power series expansion that is centred at a then,

  $f\left(x\right)=f\left(a\right)+\frac{f^{\prime }\left(a\right)\left(x-a\right)}{1!}+\frac{f^{″}\left(a\right){\left(x-a\right)}^2}{2!}+\frac{f^{‴}\left(a\right){\left(x-a\right)}^3}{3!}+\mathrm{...}$

Then,

  $f\left(x\right)=f\left(1\right)+\frac{f^{\prime }\left(1\right)\left(x-1\right)}{1!}+\frac{f^{″}\left(1\right){\left(x-1\right)}^2}{2!}+\frac{f^{‴}\left(1\right){\left(x-1\right)}^3}{3!}+\mathrm{...}$

Then, from the general equation and from the equation 1 and 2.

  $f^{\prime }\left(1\right)=-0.8$ and $f^{\prime }\left(1\right)<0$

From the graph it is seen that at $x=1$ , the graph of the function f increases $f^{\prime }\left(1\right)>0$ .

Thus, the series is not the series of f that is centred at $x=1$ as the graph of the function fincreases $f^{\prime }\left(1\right)>0$ .

(b)

To determine

To Find: The explanation for the given series no the Taylor series of $f$ ‘centred at 2.

Answer to Problem 2E

The series is not the series of f that is centred at $2$ as the graph of the function f is concave.

Explanation of Solution

Given:

The given series is,

  $2.8-0.5\left(x-2\right)+1.5{\left(x-2\right)}^2-0.1{\left(x-2\right)}^3+\mathrm{...}$ .

Calculation:

Consider that the series is the power series expansion that is centred at a then,

  $f\left(x\right)=f\left(a\right)+\frac{f^{\prime }\left(a\right)\left(x-a\right)}{1!}+\frac{f^{″}\left(a\right){\left(x-a\right)}^2}{2!}+\frac{f^{‴}\left(a\right){\left(x-a\right)}^3}{3!}+\mathrm{...}$

Then,

  $f\left(x\right)=f\left(2\right)+\frac{f^{\prime }\left(2\right)\left(x-2\right)}{1!}+\frac{f^{″}\left(2\right){\left(x-2\right)}^2}{2!}+\frac{f^{‴}\left(2\right){\left(x-2\right)}^3}{3!}+\mathrm{...}$

Then, from the general equation and from the equation above.

  $\begin{array}{l}\frac{f^{″}\left(2\right)}{2!}=1.5\\ f^{″}\left(2\right)=1.5\times 2\\ f^{″}\left(2\right)=3>0\end{array}$ and $f^{\prime }\left(1\right)<0$

From the graph it is seen that at $x=2$ , the graph of the function fis concave.

Thus, the series is not the series of f that is centred at $2$ as the graph of the function f is concave.